Answer
$${\pi ^3} - 6\pi $$
Work Step by Step
$$\eqalign{
& {\text{The travel of the particle is given by}} \cr
& s = \int_0^\pi {{t^3}\sin t} dt \cr
& {\text{Integrate by tabulation we obtain}} \cr
& s = \left[ { - {t^3}\cos t + 3{t^2}\sin t + 6t\cos t - 3\sin t} \right]_0^\pi \cr
& {\text{Evaluating}} \cr
& = \left[ { - {\pi ^3}\cos \pi + 3{\pi ^2}\sin \pi + 6\pi \cos \pi - 3\sin \pi } \right] - \left[ 0 \right] \cr
& = - {\pi ^3}\left( { - 1} \right) + 3{\pi ^2}\left( 0 \right) + 6\pi \left( { - 1} \right) - 3\left( 0 \right) \cr
& = {\pi ^3} - 6\pi \cr} $$
